When I run R3C4=8, I get different results in the three cells you mention.

Keith, I stand by my solution. An 8 in r3c4 yields a 7 in r3c7 and 2 in r3c3.

Quote:

I don't like your solution, if you don;t mind me saying so.

I don't mind you saying so at all. And if my statement above is incorrect, I'd want to know about it. I always want input if it's not an insult and has the potential to improve my game.

I'd have probably been happier with some other solution, but I have to do what I have to do in order to solve puzzles.

But just for the sake of argument, a couple of years ago I asked you about a UR you used, claiming that it looked like a Forcing Chain which you were led to by a potential UR. You replied that you saw nothing wrong with spotting a pattern and then examining the ramifications of that pattern. Is that or is that not what I did?

At any rate, I'll say the same thing you said, that I hope you don't mind me saying what I said. I'm heading to bed but look forward to your reply Saturday.

There's nothing wrong with this solution, I just don't like it. And, the longer the implication chains, the more I dislike it.

Sure, you can justify looking at whether R3C4=8 because of the UR, or because of the almost pair 35 in C4, but remember: Every bivalue cell is a finned single, or an "almost" single. Yet I would regard an almost single as simple trial and error.

If you have a Type 3, say, AB-AB-ABC-ABD, do you not look at the implications of C and D being true? How is that different from the UR here?

By the way, I also found that r3c4=8 leads to an invalidity. But here I've fallen under the influence of Ronk who has posited that common outcomes are more satisfying than invalidities. Thus, I didn't use the invalidity, rather I extended the chain to achieve a common outcome.

I agree that shorter chains are more satisfying than longer ones, but I had no choice. All the other posters are using ALS's or various Eureka-notated solutions which I don't understand.